JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

Let

\alpha

and

\beta

be the roots of

{x^2} - 2x\cos \emptyset + 1 = 0

. Then the equation whose roots are

{\alpha ^n},\;{\beta ^n}

View answer and explanation

Correct option: B

{x^2} - 2x\cos n\emptyset + 1 = 0

Option B (

{x^2} - 2x\cos n\emptyset + 1 = 0

) is correct because the roots

{\alpha ^n}

and

{\beta ^n}

satisfy this equation, derived using De Moivre's theorem.

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